Mixed anticonnected blocks lift pattern obstructions to the ambient graph

For each $i\in [s]$, apply [L1] to choose nonadjacent vertices $b_i,b_i^{\mathrm{\prime }}\in B_i$ with $v{b}_i\in E(G)$ and $v{b}_i^{\mathrm{\prime }}\mathrm{\notin }E(G)$.

Now assume $J[\{1,\dots ,q\}]=F$ and $(q-1)\eta <1$. Choose $x_1\in B_1$ arbitrarily. Suppose $x_1,\dots ,x_{j-1}$ have been chosen with $2\le j\le q$, so that for all $1\le i<k<j$ one has $x_i{x}_k\in E(G)$ if and only if $ik\in E(F)$.

Assume instead that $J[\{1,\dots ,q\}]=H$, that $12\in E(H)$, and that $(q-1)\eta <\frac{1}{2}$. Because $0<\mathrm{\mid }{N}_G(v)\cap B_i\mathrm{\mid }$ for $i=1,2$, choose $x_1\in B_1$ and $x_2\in B_2$ adjacent to $v$. Since $12\in E(H)=E(J[\{1,\dots ,q\}])$, the pair $(B_1,B_2)$ is complete, so $x_1{x}_2\in E(G)$. Suppose now that $x_1,\dots ,x_{j-1}$ have been chosen with $3\le j\le q$ so that $v{x}_1,v{x}_2\in E(G)$, $v{x}_i\mathrm{\notin }E(G)$ for $3\le i<j$, and $x_i{x}_k\in E(G)$ if and only if $ik\in E(H)$ for all $1\le i<k<j$.

Let $X=\{i_1,\dots ,i_t\}$ be a clique in $J$. By definition of $J$, the pairs $(B_{i_r},B_{i_u})$ are complete for all $r\mathrm{\ne }u$, so every vertex chosen from one selected block is adjacent to every vertex chosen from another selected block. Together with step 1.1, this shows that on the vertex set $\{v,b_{i_1},b_{i_1}^{\mathrm{\prime }},\dots ,b_{i_t},b_{i_t}^{\mathrm{\prime }}\}$ the only nonedges are $v{b}_{i_r}^{\mathrm{\prime }}$ and $b_{i_r}{b}_{i_r}^{\mathrm{\prime }}$ for $r\in [t]$. That is exactly the nonedge pattern of the complement of the $1$-subdivision of $K_{1,t}$, with $v$ as the complemented center, $b_{i_r}^{\mathrm{\prime }}$ as the subdivision vertex, and $b_{i_r}$ as the corresponding leaf.

For each $i<j$ with $ij\mathrm{\notin }E(F)=E(J[\{1,\dots ,q\}])$, the pair $(B_i,B_j)$ is not complete, so hypothesis 3 and [L3] imply that $x_i$ has at most $\eta \mathrm{\mid }{B}_j\mathrm{\mid }$ neighbours in $B_j$. Therefore at most $(j-1)\eta \mathrm{\mid }{B}_j\mathrm{\mid }$ vertices of $B_j$ violate one of the required nonadjacency conditions to the previously chosen vertices. Since $(q-1)\eta <1$ and $j-1\le q-1$, some vertex $x_j\in B_j$ avoids all those forbidden sets. For such a choice, every required edge holds automatically because whenever $ij\in E(F)=E(J)$ the pair $(B_i,B_j)$ is complete.

By induction on $j$, steps 1.2 and 2.2 produce vertices $x_1,\dots ,x_q$ with $x_i{x}_j\in E(G)$ if and only if $ij\in E(F)$ for all distinct $i,j\in [q]$. Hence $G[\{x_1,\dots ,x_q\}]$ is an induced copy of $F$. This proves assertion 2.

For $j\ge 3$, hypothesis 2 gives fewer than $\frac{1}{2}\mathrm{\mid }{B}_j\mathrm{\mid }$ neighbours of $v$ in $B_j$, so more than $\frac{1}{2}\mathrm{\mid }{B}_j\mathrm{\mid }$ vertices of $B_j$ are nonadjacent to $v$. As in step 2.2, the nonedge requirements to the previously chosen $x_i$ exclude at most $(j-1)\eta \mathrm{\mid }{B}_j\mathrm{\mid }<(q-1)\eta \mathrm{\mid }{B}_j\mathrm{\mid }<\frac{1}{2}\mathrm{\mid }{B}_j\mathrm{\mid }$ further vertices. Hence some $x_j\in B_j$ is simultaneously nonadjacent to $v$ and satisfies $x_i{x}_j\in E(G)$ if and only if $ij\in E(H)$ for every $i<j$. Inducting on $j$ produces vertices $x_1,\dots ,x_q$ such that the old vertices induce $H$, the new vertex $v$ is adjacent exactly to $x_1$ and $x_2$, and therefore $G[\{v,x_1,\dots ,x_q\}]$ is an induced copy of $H^{+}$ by [L4]. This proves assertion 3.